📝 题目
例 8 设 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上有二阶导数, $\left| {f\left( x\right) }\right| \leq a,\left| {{f}^{\prime \prime }\left( x\right) }\right| \leq b$ , 其中 $a,b$ 是非负数. 求证: 对一切 $c \in \left( {0,1}\right)$ 有 $\left| {{f}^{\prime }\left( c\right) }\right| \leq {2a} + \frac{b}{2}$ .
思路 本题条件和本题结论之间的联系是要从函数和二阶导数的估计导出一阶导数的估计. 能将函数、一阶导数和二阶导数全部联系在一起的数学工具惟有泰勒公式. 故应从泰勒公式入手.
💡 答案与解析
证 对任意给定的 $c \in \left( {0,1}\right)$ ,因为函数 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上有二阶导数,所以函数 $f\left( x\right)$ 在点 $x = c$ 处二阶泰勒公式成立,即
$$ f\left( x\right) = f\left( c\right) + {f}^{\prime }\left( c\right) \left( {x - c}\right) + \frac{{f}^{\prime \prime }\left( \xi \right) }{2}{\left( x - c\right) }^{2}\;\forall x \in \left\lbrack {0,1}\right\rbrack , $$
(5.4)
其中 $\xi$ 在 $c$ 与 $x$ 之间. 特别地,当 $x = 0$ 与 $x = 1$ 时,(5.4) 式给出
$$ f\left( 0\right) = f\left( c\right) + {f}^{\prime }\left( c\right) \left( {0 - c}\right) + \frac{{f}^{\prime \prime }\left( {\xi }_{0}\right) }{2}{\left( 0 - c\right) }^{2}\;\left( {0 < {\xi }_{0} < c}\right) ; $$
(5.5)
$$ f\left( 1\right) = f\left( c\right) + {f}^{\prime }\left( c\right) \left( {1 - c}\right) + \frac{{f}^{\prime \prime }\left( {\xi }_{1}\right) }{2}{\left( 1 - c\right) }^{2}\;\left( {c < {\xi }_{1} < 1}\right) . $$
(5.6)
由 (5.6) 减去 (5.5) 式,得到
$$ f\left( 1\right) - f\left( 0\right) = {f}^{\prime }\left( c\right) + \frac{1}{2}\left\lbrack {{f}^{\prime \prime }\left( {\xi }_{1}\right) {\left( 1 - c\right) }^{2} + {f}^{\prime \prime }\left( {\xi }_{0}\right) {c}^{2}}\right\rbrack , $$
即
$$ {f}^{\prime }\left( c\right) = f\left( 1\right) - f\left( 0\right) - \frac{1}{2}\left\lbrack {{f}^{\prime \prime }\left( {\xi }_{1}\right) {\left( 1 - c\right) }^{2} + {f}^{\prime \prime }\left( {\xi }_{0}\right) {c}^{2}}\right\rbrack . $$
再由绝对值不等式, 得
$$ \left| {{f}^{\prime }\left( c\right) }\right| \leq \left| {f\left( 1\right) }\right| + \left| {f\left( 0\right) }\right| + \frac{1}{2}\left\lbrack {\left| {{f}^{\prime \prime }\left( {\xi }_{1}\right) }\right| {\left( 1 - c\right) }^{2} + \left| {{f}^{\prime \prime }\left( {\xi }_{0}\right) }\right| {c}^{2}}\right\rbrack $$
$$ \leq {2a} + \frac{b}{2}\left\lbrack {{\left( 1 - c\right) }^{2} + {c}^{2}}\right\rbrack \leq {2a} + \frac{b}{2}\left\lbrack {\left( {1 - c}\right) + c}\right\rbrack $$
$$ = {2a} + \frac{1}{2}b\text{ . } $$